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Katena32 [7]
3 years ago
13

On what factors capacitance of parallel plate capacitor depends?

Physics
1 answer:
Roman55 [17]3 years ago
3 0

Answer:

Separation between the plates, area of the plates and dielectric constant

Explanation:

The capacitance of a parallel plate capacitor is given by:

C=k \epsilon_0 \frac{A}{d}

where

k is the dielectric constant

\epsilon_0 is the vacuum permittivity (which has a constant value)

A is the area of the plates

d is the separation between the plates

Therefore from the formula we see that the capacitance of a parallel plate capacitor depends on the following factors:

- Separation between the plates

- Area of the plates

- Dielectric constant

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The velocity is v_b = 20.17 \ m/s

Explanation:

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   The mass of the ball is  m = 0.245 \ kg

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   The force is  F =  30.9 \ N

   The speed of the ball is  v = 16.0 \ m/s.

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    K_t  =  \frac{1}{2} *  m  *  v^2

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=> K_t  =  31.36 \ J  

Generally the work done by the force applied on the ball from the top to the bottom  is mathematically represented as

       W =  F *  d

Here  d is the length of  a semi - circular arc which is mathematically represented as

       d =  \pi * r

So

      W =  30.9 *  0.598

      W = 18.48 \ J

Generally the kinetic energy at the bottom is mathematically represented as

      K_b  =  \frac{1}{2} *  m *  v_b^2

=>    K_b  =  \frac{1}{2} *  0.245  *  v_b^2

=>   K_b  =  0.1225  *  v_b^2

From the law of energy conservation

     K_t +  W  =K_b

=>    31.36+  18.48 = 0.1225  *  v_b^2

=>    v_b = 20.17 \ m/s

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